PROBABILITY
A card is drawn at random from a normal 52 -card pack and its identity noted. The card
is replaced, the pack shuffled and the process repeated. Random variablesW, X, Y, Zare
defined as follows:W=2 if the drawn card is a heart;W=0otherwise.
X=4 if the drawn card is an ace, king, or queen;X=2if the card is
ajackorten;X=0otherwise.
Y=1 if the drawn card is red;Y=0otherwise.
Z=2 if the drawn card is black and an ace, king or queen;Z=0
otherwise.Establish the correlation matrix forW, X, Y, Z.The means of the variables are given by
μW=2×^14 =^12 ,μX=(
4 × 133
)
+
(
2 × 132
)
=^1613 ,
μY=1×^12 =^12 ,μZ=2× 526 = 133.The variances, calculated fromσ^2 U=V[U]=E
[
U^2
]
−(E[U])^2 ,whereU=W,X,Yor
Z,are
σ^2 W=(
4 ×^14
)
−
( 1
2) 2
=^34 ,σ^2 X=(
16 × 133
)
+
(
4 × 132
)
−
( 16
13) 2
=^472169 ,
σ^2 Y=(
1 ×^12
)
−
( 1
2) 2
=^14 ,σ^2 Z=(
4 × 526
)
−
( 3
13) 2
= 16969.
The covariances are found by first calculatingE[WX] etc. and then formingE[WX]−μWμX
etc.
E[WX]=2( 4 )( 3
52)
+2( 2 )
( 2
52)
= 138 , Cov[W,X]= 138 −^12( 16
13)
=0,
E[WY] = 2(1)
( 1
4)
=^12 , Cov[W,Y]=^12 −^12( 1
2)
=^14 ,
E[WZ]=0, Cov[W,Z]=0−^12( 3
13)
=− 263 ,
E[XY] = 4(1)
( 6
52)
+2(1)
( 4
52)
= 138 , Cov[X, Y]= 138 −^1613( 1
2)
=0,
E[XZ] = 4(2)
( 6
52)
=^1213 , Cov[X, Z]=^1213 −^1613( 3
13)
=^108169 ,
E[YZ]=0, Cov[Y,Z]=0−^12( 3
13)
=− 263.
The correlations Corr[W,X] and Corr[X, Y] are clearly zero; the remainder are given by
Corr[W,Y]=^14( 3
4 ×
1
4)− 1 / 2
=0. 577 ,
Corr[W,Z]=− 263( 3
4 ×
69
169)− 1 / 2
=− 0. 209 ,
Corr[X, Z]=^108169( 472
169 ×
69
169)− 1 / 2
=0. 598 ,
Corr[Y,Z]=− 263( 1
4 ×
69
169)− 1 / 2
=− 0. 361.
Finally, then, we can write down the correlation matrix:
ρ=